Reliability
Last week's question.
Mean Time Between Failures (MTBF)
An electronic assembly consists of three subassemblies with the following individual MTBF.
Subassembly 1 547 days
Subassembly 2 180 days
Subassembly 3 276 days
Failures occur at random and if any subassembly fails, the entire assembly fails. What is the MTBF of the electronic assembly?
a) 334 days
b) 180 days
c) 91 days
d) None of the Above
Answer
c) 91 days
When failures occur at random, the time between failures follow an exponential distribution. If the individual subassemblies fail at random, then the entire electronic assembly will fail at random. The assembly failure rate (λ) is equal to the sum of the subassembly failure rates. Since failure rate is the reciprocal of MTBF, the assembly MTBF may be calculated as follows.
MTBF = 1/Σλi = 1/(1/547 + 1/180 + 1/276) = 91 days
Source: Carr, Wendell E., Statistical Problem Solving, ASQC Quality Press, Milwaukee WI, 1992
This week's question.
How Much to Stock?
History shows that a critical part for a manufacturing tool fails at random, with a constant failure rate of 0.35 failures per month. Spare parts are ordered once a month. How many parts should you have on the shelf at the start of a month to be at least 99% confident that you’ll have enough parts to make it through the month?
a) 1
b) 2
c) 3
d) None of the Above
Answer
b) 2
Let x be the variable number of failures in a month, and P(x or less) be the probability of x or less failures. We want to find x such that P(x or less) ≥ 0.99. Since the failure rate is constant, x follows a Poisson distribution with failure rate λ = 0.35. From the equation of Poisson probabilities below, we see P(1 or less) = 0.951 and P(2 or less) = 0.994. Thus, 2 parts is the required stock level.
P(x) = [λ^x * e^-λ]/x!
Lambda 0.35
x P(x) P(1 or 0) P(2, 1 or 0)
0 0.704688 0.951 0.994
1 0.246641
2 0.043162
3 0.005036
4 0.000441
5 3.08E-05
Source: Carr, Wendell E., Statistical Problem Solving, ASQC Quality Press, Milwaukee WI, 1992